The actual problem of the week #3
Well of course I wasn't being serious with expecting people to solve one of those millenium problems in the space of one week, so here is my actual problem of the week 3, this time a probability problem (a wonderful mathematical skill that is sadly lacking in most people). Although I'll admit, that in the case of this particular problem, figuring out the solution is mostly a problem in algebra.
The problem goes like this:
You have a bag (which you can't see inside of) that contains r red marbles and b black marbles, identical in every way except for colour (or color to you americans). You reach in, randomly pull out a marble, and then without replacement, reach in again and randomly pull out another marble. If the probability that you pull out two red marbles is exactly 0.5, what are some possible values for r and b? (i.e. the number of red and black marbles originally in the bag)
And yes, for this question both r and b must be positive integers.
There is more than one answer. In fact there are many answers. Full points if you can find a general way (or formula) of expressing all the answers.
And well done to Sherif for solving problem of the week 2.
The problem goes like this:
You have a bag (which you can't see inside of) that contains r red marbles and b black marbles, identical in every way except for colour (or color to you americans). You reach in, randomly pull out a marble, and then without replacement, reach in again and randomly pull out another marble. If the probability that you pull out two red marbles is exactly 0.5, what are some possible values for r and b? (i.e. the number of red and black marbles originally in the bag)
And yes, for this question both r and b must be positive integers.
There is more than one answer. In fact there are many answers. Full points if you can find a general way (or formula) of expressing all the answers.
And well done to Sherif for solving problem of the week 2.
4 Comments:
Well, one week later and no responses. I thought at least one person would have come up with the first combination just by a bit of trial and error: 3 red marbles and one black marble. (You then have a 3 in 4 chance of drawing red the first time, and if you do this you then have a 2 in 3 chance of drawing red the second time. And then 3/4 * 2/3 = one half.)
So that?s that. As I said though, there are many solutions (an infinite number in fact, but they get pretty big pretty fast). I?ll spare you the details of the proof that what?s below incorporates every possible combination, always gives integer values, and always gives a probability of one half. Just take my word for it.
About the simplest way I can express the answer is this:
R = (2*sqrt(2) + (1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1))/(4*sqrt(2))
B = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))/(4*sqrt(2))
And then for each value positive integer value of n (such as 1, 2, 3, 4,? etc) you get a different set of values for R and B. If you let n = 1, you get R = 3 and B = 1, as mentioned above. The first 5 answers are:
n: 1, 2, 3, 4, 5
R: 3, 15, 85, 493, 2871
B: 1, 6, 35, 204, 1189
Another way to express the answers (although not as neat at generating them I don?t think) is like this:
Let X0 = 0
Let X1 = 1
Let Xn = 2*X(n-1)+X(n-2) for n greater than or equal to 2.
(This same sequence can be expressed explicitly as Xn = ((1+sqrt(2))^n-(1-sqrt(2))^n)/(2*sqrt(2)) if you prefer.)
Then, the values for B are the partial sums of the Xn sequence for odd values of n, and the values for R are the partial sums of the Xn sequence for even values of n, plus 1.
(Eg, for the third solution R=85, B=35, we get R=X2+X4+X6+1, and B=X1+X3+X5.)
I?ll try to think of something a little more popular for the next problem.
Cheers.
Well, that was just hard. :/
chocolate
Just a correction, there was a typo in my original answer (I hate it when that happens). In the first formula for R, the part that says ^(2*n-1) should actually say ^(2*n+1). This correction should be made for both cases.
Shame on Roslyn and Leon for not picking up on that! You're getting slack guys.
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